OPERATION. 1 1 2 3 4 5 6 7 8 Multiplicand. 6 7 407 40 68 = Product by 6 units. 3 2 3 5 5 5 5 5 2 6 4 = 1st product x 8 tens for product by 48 tens. 2d product x 3 thousands for product by 144 thousands. 3d product X 2 millions for product by 288 millions. 3 2 3 7 1 7 8 7 6 4 1 6 3 1 5 0 8 = Product by 288144486. 3. Multiply 61370913 by 96488. Ans. 5921556653544. 4. Multiply 8649347864 by 1325769612. Ans. 11467042561708308768. 94. When the multiplier is any number of nines. Ex. 1. Multiply 87654 by 999. Ans. 87566346. By annexing three ciphers to the multiplicand we take it 1000 times, or 1 time more than is re quired by the given multiplier. We therefore from this result subtract the multiplicand taken once, and thus obtain the product by 999. In like manner we may multiply by any number of nines. Hence the RULE.- Annex as many ciphers to the multiplicand as there are nines in the multiplier, and from the number thus produced subtract the given multiplicand. NOTE. To multiply by any number of threes, find the product for the same number of nines, by the rule, and take one third of it by dividing by 3; and to multiply by any number of sixes, take twice the product of the same number of threes, by multiplying it by 2. 10. Multiply 1234567 by 9999. 11. Multiply 98123452 by 999999. Ans. 98123353876548. Ans. 12344435433. CONTRACTIONS IN DIVISION. 95. When the divisor is a convenient part of a number of tens, hundreds, or thousands. Ex. 1. Divide 19641075 by 25. OPERATION, 19641075 4 Ans. 785643. By multiplying both divisor and dividend by 4, which does not change the relation of the one to the other (Art. 83), the divisor becomes 100, 7 8 5 6 4 300 Quotient. which enables us to perform the di vision by simply cutting off two figures at the right of the dividend (Art. 79). In like manner, if the divisor had been 34, by taking both it and the dividend 3 times as large, we could have performed the division by simply cutting off one figure at the right of the dividend; or, if the divisor had been 125, by taking both it and the dividend 8 times as large, we could have performed the division by simply cutting off three figures at the right of the dividend. Hence the RULE. Multiply both divisor and dividend by that number which will change the divisor to a number of tens, hundreds, or thousands, and then divide. 9. J. Cushing bought a number of horses, at 1663 dollars each, for $9500; required the number bought. Ans. 57 horses. 10. A company has received 12000 dollars from the sale of piano-fortes, at an average price of 3334 dollars each; required the number sold. 11. How many shares of the Illinois Central Railroad, at 125 dollars each, can be bought for 150000 dollars? Ans. 1200 shares. 12. How many cows, at 334 dollars each, may be purchased for 333 dollars. 13. How many books, at 24 dollars each, may be purchased for 120 dollars? Ans. 48 books. 14. A certain magazine contains 616350 pounds of powder, in kegs of 25 pounds each; required the number of kegs. Ans. 24654 kegs. 96.° When the divisor is any number of nines. Ex. 1. Divide 316234 by 99. OPERATION. 316234 3196 127 28 28 3194 Ans. Ans. 319438. We first divide by 100, or 99+1, by cutting off two figures at the right of the dividend, obtaining for the first partial quotient 3162 and a remainder 34. Since the divisor used was 1 larger than the given divisor, 99, the quotient obtained denotes that an excess of 3162, or a number equal to the quotient itself, must be added to the 34 for the true remainder, 34 +3162 3196, which exceeding the divisor 99, we write it for a second dividend; and dividing by 100, or 99 + 1, as before, we obtain for the second partial quotient 31, and a remainder 96. To the remainder 96 we add 31, the excess denoted by the last quotient, and obtain 127 for a third dividend, which being divided by 100, or 991 gives the third partial quotient 1, and a remainder 27. To the remainder 27 adding 1, the excess denoted by the last quotient, we have for the true remainder 28, which is 38. The sum of the partial quotients with the final remainder annexed gives 31948, the quotient required. 100 = = = The above process is based upon the principle that 10 991; 1000 9991, etc.; consequently 20 +2, and 37 (3 × 9) + 37; 200 (38 x 99) + 38 +59; 15987 Hence, 316234 = 31620034 316234 3196 96 127 = = = = = 9 + 1; (2 × 9) = (2 × 99) + 2; 3859 (15999) 15+ 987, etc. (3162 × 99) + 3162 + 34; 3100 +96 (31 × 99) +31 +96; 31 + = 100+ 27- (1 × 99) +1 + 27; 1 + 27 and 316231+1+99 31943, the answer, as before obtained. By like process, and upon the same principle, may the quotient be found for any number of nines. RULE. Add 1 to the given divisor, and, by it thus increased, divide by cutting off figures at the right of the dividend. To the figures cut off on the right add those on the left for a true remainder, of which, if it equal or exceed the given divisor, make a second dividend, and divide as before. Proceed thus till there shall be no remainder as large as the given divisor; and the sum of the several quotients, with the last remainder, if any, will be the answer required. NOTE. When the last remainder is the same as the given divisor, it must be cancelled, and 1 written as a partial quotient. 97.° Abbreviated method of long division. Ex. 1. Divide 34634 by 134. OPERATION. 134) 346 34 (258 62 Ans. 783 1134 = Ans. 2586 now bring down 3, and find = 3; 5, and 1 the next quotient figure to be 5, then, 4 X 5 20; 3-0 15, and 2 carried = 17; 8-7 = 1; 1 X 5: 6 6; 7 = 1. We next bring down 4, and find the next be 8; then, 4 × 8: 3 X 5 = ried = = 27; 13 = = 32; 4 2= 2; 3 X 8 - - 7 = 6; 1 X 8 = 8, and 2+1 car1 0; 1 0. Hence, this method is that of ordinary long division (Art. 73), abridged by subtracting each figure of the product of the divisor and a quotient figure, as it is obtained, and writing down only the remainders. EXAMPLES. 2. Divide 39006 by 44. Ans. 8862. 3. A certain orchard contains 1088 trees in 34 rows; how many trees in each row? Ans. 32 trees. 4. A speculator sold 191 mules at a gain of 5157 dollars what was the gain on each? ; 5. The population of Massachusetts, in 1855, was 1,133,123; how many would that be to each of its 7750 square miles of surface? Ans. 146493. 98. When the divisor or dividend has a fraction annexed. 83692 3 3 Ans. 426. We bring the divisor and dividend to the same fractional parts as those denoted by the given 2 6 ) 1 1 0 7 6 ( 426 Ans. fraction, which are thirds, by mul 104 67 52 156 156 tiplying them both by 3, the number of thirds in 1. We thus change the divisor to 26 thirds, and the dividend to 3692 thirds, without changing the quotient (Art. 83); and dividing obtain 426, the answer required. RULE. Reduce the divisor and dividend to the same fractional parts as are denoted by the given fraction, and then divide as in whole numbers. NOTE. In case there should be a remainder after the division, its true value may be found as in Art. 76. 5. If 16 feet make a rod, how many rods are there in 10626 feet? 6. If 69 miles make a degree, how many degrees in 12450 miles? Ans. 180 degrees. 7. How many barrels of pork, at 17 dollars a barrel, may be bought for 5591 dollars? 8. A merchant has sold 2667 yards of silk in dress patterns of 10 yards each; required the number of patterns. Ans. 254 patterns. 9. How many plats, each of 2721 square feet, are equal in extent to 136125 square feet? Ans. 500 plats. 10. How many days will 119 pounds of bread last a man, if Ans. 49 days. 'he consume 2 pounds per day? 11. How many casks of 31 gallons each will be required to hold 129683 gallons of cider? Ans. 415 casks. |